Permutation Calculator
Calculate P(n,r) — the number of ordered arrangements of r items chosen from a set of n when order matters.
Permutation Calculator
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Formula: P(n,r) = n! / (n-r)!
Worked example — P(8,3) = 336 possible medal orderings
Formula
P(n,r) = n! / (n-r)!
Permutations count the number of ordered sequences of length r that can be formed from n distinct items, where each item can be used at most once. It equals n × (n-1) × (n-2) × ... × (n-r+1), which simplifies to n!/(n-r)!.
Worked Examples
Example 1: Awarding medals
Problem:In how many ways can gold, silver, and bronze medals be awarded among 8 finalists?
Solution:P(8,3) = 8! / (8-3)! = 8! / 5! = 8×7×6 = 336.
Result:P(8,3) = 336 possible medal orderings
Example 2: Password without repeated characters
Problem:How many 4-character passwords can be formed from a 10-character alphabet if no character may repeat?
Solution:P(10,4) = 10! / (10-4)! = 10×9×8×7 = 5,040.
Result:P(10,4) = 5,040 possible passwords
Frequently Asked Questions
What is a permutation in mathematics?
A permutation is an ordered arrangement of r items selected from a larger set of n items, where the sequence matters. P(n,r), read 'n permute r', counts how many distinct ordered sequences of length r can be built — assigning gold, silver, and bronze medals to 3 of 8 runners is a permutation problem because swapping which runner gets which medal creates a different outcome.
What is the difference between a permutation and a combination?
A permutation counts arrangements where order matters (1st, 2nd, 3rd place finishers), while a combination counts selections where order doesn't matter (choosing any 3 team captains with no ranking). Because every group of r items can be arranged in r! different orders, P(n,r) = C(n,r) × r! — permutations are always at least as large as the matching combination count.
How is the permutation formula derived?
To fill r ordered positions from n available items: there are n choices for the first position, n-1 remaining choices for the second (since one item is used), n-2 for the third, and so on down to n-r+1 choices for the last position. Multiplying these together, n × (n-1) × (n-2) × ... × (n-r+1), simplifies to n! / (n-r)! — the standard permutation formula.
What does P(n,n) equal, and why?
P(n,n) = n! — the total number of ways to arrange all n items in a sequence with no items left over. For example, P(5,5) = 5! = 120, matching the well-known fact that 5 distinct books can be arranged on a shelf in 120 different orders.
What real-world problems use permutations?
Permutations determine race-finish orderings (who gets 1st, 2nd, 3rd), PIN and password possibilities when digits or characters cannot repeat, seating arrangements at an event, the number of possible race schedules or tournament brackets, and route-ordering problems like the traveling salesman problem's underlying arrangement count.
How do permutations with repetition differ from Permutation Calculator's result?
Permutation Calculator computes permutations without repetition, where each of the n items can be used at most once in the arrangement (standard for people, cards, or unique objects). Permutations with repetition allowed (like a 4-digit PIN where digits can repeat) instead use nʳ, a different and typically much larger formula, since each of the r positions independently has n choices.
Why does P(n,r) grow so much faster than C(n,r) as r increases?
P(n,r) counts every distinct ordering of each selected group separately, and the number of orderings of an r-item group is r! (which grows extremely fast — 5! = 120, 10! = 3,628,800). Since P(n,r) = C(n,r) × r!, the gap between the two widens rapidly as r grows, even though both describe selections from the same underlying set.
How large can permutation results get?
Even modest inputs produce enormous numbers — P(20,10), the number of ways to award 10 distinct ordered prizes among 20 contestants, is 670,442,572,800 (over 670 billion). This rapid growth is exactly why permutations are used to argue that brute-force password guessing or exhaustive search of large arrangement spaces is computationally infeasible.
What is a circular permutation, and why is it not P(n,n)?
When n distinct items are arranged around a circle, rotating the whole arrangement does not create a new outcome, so the n! linear arrangements collapse into groups of n rotations that are all equivalent. The count is therefore (n-1)! rather than n!. Seating 6 guests around a round table gives 5! = 120 distinct arrangements, not 720, and if mirror images also count as the same the total halves again to (n-1)!/2.
How do you count arrangements when some of the items are identical?
Use the multiset permutation formula n! / (n1! x n2! x ... x nk!), where each nk is the count of one repeated item. Dividing by the factorial of each repeat cancels the orderings that swap indistinguishable copies. Arranging the 11 letters of MISSISSIPPI gives 11! / (4! x 4! x 2! x 1!) = 39,916,800 / 1,152 = 34,650 distinct strings, far fewer than the 11! you would get if every letter were unique.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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